hw7sol - Math 131A - Section 2 Spring 2010 Homework 7 The...

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Math 131A - Section 2 Spring 2010 Homework 7 The following problems were graded on this assignment: 18.4 (2 points), 18.8 (2 points), 18.12 (3 points), 19.1(b,e) (2 points), 19.4 (3 points), 19.7 (3 points) 18.4. Let f ( x ) = ( x - x 0 ) - 1 . Observe that this function is continuous for every x 6 = x 0 by 17.4, so f is continuous on S since x 0 / S . Let M > 0 and let ± = M - 1 . Then, since lim x n = x 0 , there is N such that | x n - x 0 | < ± for n > N . Then, | f ( x N +1 ) | = | x N +1 - x 0 | - 1 > M , so f is unbounded on S . 18.8. If a = b , then f ( a ) f ( b ) = f ( a ) 2 0, a contradiction. Hence, we may assume without loss of generality that a < b . Note that f ( a ) 6 = 0, since this implies f ( a ) f ( b ) = 0. Suppose f ( a ) < 0. Then, since f ( a ) f ( b ) < 0, we have that f ( b ) > 0. Then, 0 ( f ( a ) ,f ( b )), so there exists x ( a,b ) such that f ( x ) = 0 by the intermediate value theorem. Similarly, if f ( a ) > 0, then f ( b ) < 0, so there exists x ( a,b ) such that f ( x ) = 0 since 0 ( f ( b ) ,f ( a )) by the intermediate value theorem, giving the desired result. 18.12.(a). This result is shown in the last homework assignment, so a solution can be seen there.
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This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.

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